Honours Analysis III

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1 Honours Analysis III Math 354 Prof. Dmitry Jacobson Notes Taken By: R. Gibson Fall

2 Contents 1 Overview p-adic Distance Introduction Normed Linear Spaces Inner Product Spaces Metric Space Techniques Completeness Basics & Definitions Completion & Density Revisited O.D.E. & Contraction Compactness Basics & Definitions Arzela & Compacta Basic Point-Set Topology Product Topology Conectedness Connected Components/ Path Components Banach Space Techniques Linear Functionals Baire s Category Theorem, Banach-Steinhaus Theorem & The Open Mapping Theorem Convex Sets Kernel Conjugate Space Linear Functionals Revisited Bernstein Polynomials Inverse & Implicit Function Theorem In R n Linear Operators & The Operator Norm The Hahn-Banach Theorem Examples

3 Chapter 1 Overview Definition 1. (i) x X, d(x, x) = 0 (ii) x y X, d(x, y) > 0 (iii) x, y X, d(x, y) = d(y, x) Let X be a metric space. We define Distance d : X X R to satisfy (iv) x, y, z X, d(x, z) + d(z, y) d(x, y) Example 1. The following are some of the main examples of metric spaces for this course. Spaces of sequences (finite or infinite) - with l p norm. {(x 1,..., x n ),...) : This leads us to a proposition. Proposition 1. If x = (x 1, x 2,...), y = (y 1, y 2,...), then is a distance. x k p < } k=1 ( ) 1/p d(x, y) = x k y k p k=1 Example 2....probably not. Example 3. If x = 0 and y = (1, 1/2, 1/3, 1/4,...), then can we prove that d(x, y) = π h 2 + = 2 6 For p 1, an example would be continuous functions f, g on [a, b]. Proposition 2. In the space of continuous functions on [a, b], [ b d(f, g) = f(x) g(x) p dx a defines a distance. We recall the L p norm to be ( b f p = f(x) p dx a ) 1/p ] 1/p 3

4 We now examine polygons in the plane. There is a problem in the assignment which asks if the difference of areas defines a distance. So we must prove that d(p 1, P 2 ) = Area(P 1 P 2 ) which is the area of (P 1 P 2 )\(P 1 P 2 ) p-adic Distance Section 1.1 Let p {2, 3, 5, 7, 11, 13,...} = P rimes. Let x, y Q so x = a 1 b 1 y = a 2 b 2 Definition 2. where k Z and The p-adic Norm is defined as follows x = p k c d x p = p k Definition 3. The p-adic Norm is defined to be d p (x, y) = x y p Example 4. Working with p-adic distance. Suppose that then we have that x = x = 7 1 = x 7 = 7 2 x = = x 3 = 3 1 x = = x 2 = 2 3 Proposition 3. x y p defines a p-adic distance on Q. NB: If x is as above, then x Eucl x 2 x 3 x 7 = 1 Neat, eh? 4

5 Chapter 2 Introduction Normed Linear Spaces Section 2.1 Linear means exactly what you would think it means. A good way to show this is (a n ) n=1 + (b n ) n=1 = (a n + b n ) n=1 Definition 4. (f + g)(x) = f(x) + g(x) x [a, b] Suppose that X is a linear space, then we say that the Norm of x X is a map : X R + such that (i) x = 0 x = 0 (ii) t x = t x (iii) x + y x + y It is a fact that defines a distance since d(x, y) = x y d(y, x) = y x = (x y) = x y = d(x, y) d(x, z) = x z = (x y) + (y z) x y + y z = d(x, y) + d(y, z) Inner Product Spaces Section 2.2 An Inner Product Space is a space together with a map such that (i) 0 (x, x) (ii) (x, y) = 0 x = 0 (iii) (x, αy + βz) = α(x, y) + β(x, z), (, ) : X X R x, y, z X, α, β R (iv) (γx + δy, z) = γ(x, z) + δ(y, z) Example 5. The following are some examples of inner products 5

6 Dot Product (x, y) = x y = x 1 y x n y n Function Space Inner Product and f, g C([a, b]) = (f, g) = b b a f(x)g(x)dx f = a [f(x)] 2 dx Proposition 4. always defines a norm. We will need a lemma. x := (x, x) Lemma 5. The following identity holds. (x, y) x y we know that now, (x + ty, x + ty) = (x, x) + 2t(x, y) + t 2 (y, y) x + ty 2 0 D = 4(x, y) 2 4(x, x) (y, y) 0 (x, y) (x, x) (y, y) = x y With this lemma, we can now prove the above proposition. So The triangle inequality states that x + y x + y and now we can square both sides to obtain and now and by the lemma, and we cancel to obtain ( x + y ) 2 ( x + y ) 2 (x + y, x + y) = (x, x) + 2(x, y) + (y, y) x x y + y 2 = (x, x) + 2 x y + (y, y) which yields our result. How to guess whether x comes from (x, x) or not? The answer is = (x, y) x y (x + y, x + y) + (x y, x y) = (x, x) + 2(x, y) + (y, y) + (x, x) 2(x, y) + (y, y) 6

7 then ( ) x + y 2 + x y 2 = 2 x y 2 The norm comes from (, ) ( ) holds for all x, y X. Proposition 6. Let (a 1,..., a n,...) be such that a 2 i < and let (b 1,..., b n,...) be such that b 2 j < and let Then and so as n. To prove convergence, a = ( i=1 a 2 i ) 1/2 a i b i a b i=1 (a, b) = a 1 b a n b n n a 2 i n n+m k=n i=1 a i b i a 2 k n+m k=n k=n b 2 k b 2 j a b the two components on the right converge separately to 0 as n. This is because a 2 k < k=1 and the same goes for the sum of b 2 k. Thus, the whole thing converges to 0. Summarizing, we proved that (C([a, b]), L 2 -norm) is a metric space, and so is l 2. Next, we look at L p -norm which is and at l p -space, p 2, we have Thus, l p is the space of all a such that To prove this we need a lemma. [ b f p = f(x) p dx a ] (1/p) ( ) (1/p) a lp = a k p k=1 a k p < k=1 7

8 Lemma 7. Let a > 0, b > 0, and p, q 1 and such that 1 p + 1 q = 1 NB: We say that p, q are Conjugate Exponents. Then ( ) a b ap p + bq q Let x > 0, then f(x) = log (x) which is a concave function. f (x) = 1 x f (x) = 1 x 2 < 0 and since f is concave, we can say that f(αx + βy) αf(x) + βf(y) where α + β = 1. Also, we know that log (a b) = log a + log b ( ) = 1 p log (ap ) + 1 q log (bq ) ( 1 log p ap + 1 ) q bq Theorem 8 (Holder s Inequality). Let p < 1 where Remark. and so we say that ( n n ) 1/p ( n ) 1/q a k b k a k p b k q k=1 k=1 1 p + 1 q = 1 Both sides are homogeneous of degree 1 in (a k ), (b k ). So WLOG, we can rescale a k s so that k=1 n a k p = 1 k=1 n b k q = 1 k=1 k=1 n n ( ak p a k b k + b k q ) p q k=1 ( = 1 n ) ( a k p + 1 n ) b k q p q k=1 = 1 p + 1 q = 1 k=1 8

9 by our normalization of the sums. Now, and so ( n ) 1/p ( n ) 1 q RHS = a k p b k q k=1 k=1 LHS RHS = 1 1/p 1 1/q = 1 Now, if we let n, then we get Holder for infinite series. The proof is left to the reader as an excercise. Theorem 9 (Minkowski Inequality). (For Sequences With n < ) Recall that for x = (x 1,..., x n ), we have We want So, we know that And now, and so RHS = x + y p p = = n x k + y k p k=1 n ( x k + y k ) p k=1 ( n ) 1/p x p = x k p k=1 x + y p x p + y p n [( x k + y k ) p 1 x k + ( x k + y k ) p 1 y k ] k=1 and we apply Holder to each inner sum and let b k = ( x k + y k ), a k = x k ( ( k = 1 n x k p) 1/p n ) 1/q [( x k + y k ) p 1 ] q k=1 ( ( + k = 1 n y k p) 1/p n ) 1/q [( x k + y k ) p 1 ] q k=1 1 p + 1 q = 1 = 1 1 p = 1 q = p 1 = 1 p q = (p 1)q = p [ ( k = 1 n x k p) 1/p ( + k = 1 n y k p) ] 1/p [ k = 1 n ( x k + y k ) q] 1/q and n RHS LHS = ( x k + y k ) p k=1 [ n ] 1 1/q = ( x k + y k ) p x p + y p k=1 9

10 and we can conclude that x + y p x p + y p Example 6. The unit sphere in { } n l p (R n ) = x R n : x k p = 1 k=1 For n = 2, p = 1, we have For n = 2, p = 2, we have Now we ask what happens when p? We get {(x, y) : x + y = 1} {(x, y) : x 2 + y 2 = 1} ( [ x k p + y k p ] 1/p = ( x k p ) 1/p 1 + as p. Now as another example, we have as p. ( ) p ) 1/p yk x x k (x, y) p max{ x, y } = (x, y) Definition 5. The l norm of x = (x 1,..., x n ) is x = and for f C([a, b]), we get that the l norm of f is n max k=1 {x k} f = max x [a,b] f(x) and we can show that as p, for any f C([a, b]). f p f Metric Space Techniques Section 2.3 Definition 6. We say that (X, d) is a Metric Space if d(, ) defines a distance on the set X. Definition 7. Let A X where X is a metric space. Let x X (may or may not be in A). If every ball B(x, r) centred at x of radius r has at least one point from A for any r > 0, then this is equivalent to calling x a Contact Point. Also, just for notational purposes, B(x, r) = {y X : d(x, y) < r} Remark: Any x A is a contact point of A. 10

11 Definition 8. Point. Definition 9. If B(x, r) for any r > 0 has infinitely many points from A, then we say that x is a Limit If x A, r > 0 s.t. B(x, r) A = {x} Proposition 10. x is an isolated point x A x is a limit point of A, x A x is a limit point of A, x / A Let x X be a contact point. Then, x must be one of the following. Also, x is not an isolated point if and only if (x n ) n=1 A where the x n are distinct such that x n x as n and thus lim n d(x, x n) = 0 Definition 10. The Closure A of A is the set of all contact points. A = A {Limit points of A} Proposition 11. A = A Let x A and let r > 0. We know that y A such that y B(x 1, r 1 ) B(x, r) So, x is a contact point of A, and thus x A. Proposition 12. (i) A 1 A 2 = A 1 A 2 (ii) A = A 1 A 2 = A = A 1 A 2 Definition 11. A, B X. We say that A is Dense in B if B A and it is also true that A is dense if and only if A is dense in X. Example 7. Points with rational coordinates are dense in R k. Definition 12. Example 8. distance. We say that X is Separable if X has a countable, dense subset. Let X be the set of all bounded sequences of real numbers. Distance on X can be the l d(x, y) = sup x n y n n (x 1,..., x n,...) = sup n x n n 11

12 x = (0.9, 0.99,..., ,...) ( 9 x = sup n ) 10 n = 1 Proposition 13. X with l distance is not separable. Look at A X, A = { all infinite sequences of 0 s and 1 s}. A is not countable by the Cantor Diagonalization Argument. Suppose it is a 1 = (ɛ 1 1,... ɛ k 1,...) a 2 = (ɛ 1 2,... ɛ k 2,...) a 3 = (ɛ 1 3,... ɛ k 3,...) b = (b 1,..., b n,...) If ɛ 1 1 = 1 then b 1 = 0. If ɛ 1 1 = 0 then b 1 = 1. If ɛ 2 2 = 0, then b 2 = 1. If ɛ 2 2 = 1, then b 2 = 0. Claim: Sequence b is different from all a i. This is a contradiction which shows that A is uncountable. A has cardinality of the continuum. Claim: If x 1, x 2 A, then d (x 1, x 2 ) = 1. Lemma 14. A is not separable. x A, y B s.t. y B(x, 1/3) Suppose that B is a countable dense subset of A. Claim: If x 1 x 2, A, y 1, y 2 B(x 1, 1/3), then y 1 y 2. This is a contradiction! Let A X be uncountable. Let x, y A, x y and d(x, y) 1. We claim that if B X is dense in X, then B cannot be countable. Let (x α ) α A be our set. Consider {B(x, 1/3) : x A} The set B is dense in X, so α A, B(x α, 1/3) contains a point y α B. Lemma 15. If x α x β, then B(x α, 1/3) B(x β, 1/3) =. We know that y α B(x α, 1/3) and y β B(x β, 1/3). Moreover, y α y β. It follows that there is a bijection between A and a subset of B. Examples Of Countable Dense Sets In C([a, b]) With Various Distances Suppose that f C k ([a, b]). The first idea is to approximate by polynomials. Bernstein polynomials approximate well. P n (x) = a n x n + + a 0 where a j R and Q n (x) = b n x n + + b 0 where b j Q. Those polynomials are dense in C([a, b]) with both L p and L. [ b d p (f, g) = f(x) g(x) p dx a d (f, g) = sup f(x) g(x) x [a,b] ] 1/p 12

13 Definition 13. N a k sin (kx) + b k cos (kx) k= N The distance from a point x to a set A is equivalent to d(x, A) = inf d(x, a) a A For example, d(x, A) = 0 if and only if x is a contact point of A. d(a, B) = inf {d(x, y)} x A,y B Another example is that if A B 0, then we can take x = y A B. Hence d(a, B) = 0. It s not true both ways. We can have A B 0 but d(a, B) = 0. Definition 14. Let A X. We say that A is Closed if A = A. Example 9. [a, b] R Some examples of closed sets are {y X : d(y, x 0 ) R > 0} or {f C([a, b]) : f(x) R} for all x [a, b]. If f(x) = 0 on [a, b], then and d(0, g) = sup g(x) x [a,b] {g : d(0, g) R} which is exactly the second item on the list above. Proposition 16. Let (A α ) α I be a collection of closed sets. Then B = α I A α is also closed. B = α A α = B = B {Limit Points of B} And B = B every limit point x of B belongs to B. So, suppose that x is a limit point of B = α A α. Let r > 0, then B(x, r) contains infinitely many points of B = α A α If y B, then y A α for all α. This means that x is a limit point of A α for all α I. A α is closed, and so every limit point is contained in A α. Thus, x A α for all α I. = x α I A α = B 13

14 Proposition 17. is closed. Let A 1,..., A n be closed. Then n B = i=1 A i B = Let x B. We shall show that x cannot be a limit point of B. if x / n i=1 n i=1 then x / A i for all i = {1,..., n}. All the A i are closed, so x cannot be a limit point of A i for any i. This implies that r > 0 such that x / A. Let r = min k d(x, y k ) then B(x, r) A = for any 1 i n. Now r i such that Let then for each i. Thus, x is not a limit point of A i A i B(x, r i ) A i = r = min 1 i n r i B(x, r) A i = n i=1 A i Definition 15. We say that x is an Interior Point of A if and only if r > 0 s.t. B(x, r) A Definition 16. We say that A is Open if every point in A is an interior point. Proposition 18. A is open if and only if X\A is closed. Suppose that A is open, and that x A, then B(x, r) A and B(x, r) A C = Hence, x is not a contact point of A C. So, (A C ) A C = A C is closed. If A C is closed, then x A = x is not a contact point of A C. This means that r > 0 such that B(x, r) A C = = x is an interior point of A. = B(x, r) A 14

15 Proposition 19. If A α is open for any α I, then B = α I A α is also open. A α is open and so equivalently, we have that A C α is closed ( α I A α)c = α I A C α is closed. So, B C is closed which is equivalent to B being open. Now we invoke Proposition 17. Proposition 20. A finite intersection of open sets is also open. That is, if A i is open for i = 0,..., n, then n i=1 is also open. Beware, however, that this is not necessarily true for the infinite case. An example of this would be that if ( A n = 1 n, ) R n A i which is clearly open for any n, then which is closed. B = A n = [0, 1] n=1 Definition 17. A collection A α, for α I of open sets is called a Basis (of all open sets) if and only if any open set in X is a union of a sub-collection of A α. Definition 18. X. X is called Second Countable if and only if there is a countable basis of open sets of Lemma 21. that {G α } forms a basis if and only if for any open set A, and for any x A, there exists α such x G α A Proposition 22. Let X be a metric space. We claim that X is second countable if and only if X is separable. Idea is to let {y 1, y 1,..., y n,...} be countable, dense subset of X. Then, {B(y j, r j ) : r j Q} is a basis of all open sets in X. (= ) Let G 1,..., G n be a countable basis. Choose x n G n. We claim that the set {x n } is dense in X. To prove this, we let x X, r > 0 and we consider B(x, r) which is an open set. Now, by the lemma above, we know that there exists m such that x G m B(x, r). It follows that x m G m B(x, r). ( =) Let {x n } be a countable dense subset of X. It suffices to show that { B ( x n, 1 k ) } : n, k N 15

16 forms a countable basis. To show this, we let A be an open subset of X and we pick x A. Choose m > 0, such that ( B x, 1 ) A m Next, we choose k such that We now claim that ( x B x k, d(x, x k ) < 1 3m ) 1 B 2m ( x, 1 m 1 2m + 1 3m = 5 6m < 1 m This claim implies the previous claim by the lemma. ) A Fact:, X are both open and closed. By X, we mean the entire metric space. Definition 19. We say that X is Connected if and only if any subset A X that is both open and closed is either or X. Example 10. R is connected, but R {0} is not. Definition 20. Let d 1, d 2 be two distances on X. We say that d 1 and d 2 are Equivalent precisely if there are two constants 0 < c 1 < c 2 < such that This implies that for r 2 < r 1 < r 3, we have c 1 < d 1(x, y) d 2 (x, y) < c 2 B d2 (x, r 2 ) B d1 (x, r 1 ) B d2 (x, r 3 ) EXCERCISE: Express r 1, r 3 if we know c 1, c 2. Finally, if d 1, d 2 are equivalent, then they define the same open and closed sets. These constants are r 1 = r, r 2 = r C, r 3 = r C and for C > 1. Corollary 23. Corollary 24. same. c 1 = 1 C, c 2 = C Open sets with respect to d 1, d 2 are the same. The topologies (space X together with a collection of open sets) defined by d 1, d 2 are the Definition 21. Let X be a set and let {A α } α I be a collection of open sets. A Topological Space satisfies (i), X are both open. (ii) is open. α J A α 16

17 (iii) is open. Proposition 25. for p 1 are equivalent. Distances in R n defined by n i=1 A i n d p (x, y) = x i y i p d (x, y) = max 1 j n x i y i 1/p Definition 22 (More General Definition). We say that x is a Contact Point of A X where X is a topological space if every neighbourhood of x contains a point in A. A sequence (x n ) n=1 x if and only if for any neighbourhood U of x, there exists N > 0 such that x n U for n N. Metric Spaces are Hausdorfff. Definition 23. We say that a space X is Hausdorff if for any x, y X such that x y, there exists r 1, r 2 such that B(x, r 1 ) B(y, r 2 ) = OR For any x y, there exists open sets U containing x and V containing y such that U V = Definition 24. We say that a topological space X is Metrizable if there exists a metric d on X such that open sets defined by d give the same topology on X. Corollary 26. Definition 25. If a topological space is not Hausdorff, then it is not metrizable. Let f : X Y. We say that f is Continuous at x X, if ɛ > 0, δ > 0 s.t. y X, d X (x, y) < δ = d Y (f(x), f(y)) < ɛ and we say that a f is a Continuous Function if it is continuous x X. OR (x n ) n=1 x ( lim n x n = x), lim n f(x n) = f(x) Proposition 27. f : X Y is continuous at every point x X if and only if for every open set U in Y, f 1 (U) X is also open. Moreover, this is also equivalent to saying that for every closed set B in Y, f 1 (B) X 17

18 is also closed. Another way to say this is X f 1 (B) = f 1 (Y B) (= ) Let B Y be closed and let f : X Y be continuous. Now let f 1 (B) = A X. We want A to be closed. It suffices to show that all limit points of A lie in A. We know that there exists (x n ) A such that x n x as n, and that f is continuous at x. By the definition of continuity, we have that f(x n ) f(x) Now, f(x n ) B, and B is closed. Thus f(x) B, but then x f 1 (B) = A. ( =) Let x X, y = f(x) and let U be an open set, and y U. Then Y U is closed. Therefore, f 1 (Y U) = f 1 (A) is closed. Moreover, x / A. So there exists an open set V where x V X A. Therefore, f(v ) U. For any sequence x n x, x n V for n N. Then f(x n ) U, and so f(x n ) f(x) as n Claim: Let K be the Cantor set. We claim that K is uncountable. The idea for the proof is that points in K are like real numbers with only 0 s and 2 s in expansion (base 3). If x [0, 1], then for a Decimal Expansion, we have x = a a n 10 n + for a j {0, 1..., 9}. We can also use base 2 for a Binary Expansion x = a a n 2 n + for a j {0, 1}. For the Cantor set, we would use a Ternary Expansion x = a a n 3 n + for a j {0, 1, 2}. Proposition 28. Points in K are in one to one correspondence with such that a j {0, 2}. The cantor set K is self similar. To show this we define a map f : K [ 0, 1 ] K 3 where f(x) = 3x. If we look at a ternary expansion in decimal notation (i.e ), then multiplying the ternary expansion of a number is just the same as multiplying a decimal expansion by 10. We now examine how to prove that a set A X is dense in B X. Assume for simplicity that A B. We need to show that x B, ɛ > 0, y B(x, ɛ) A a j 3 j 18

19 Now, if we construct the cantor set by letting [0, 1] = I 0 1 and letting the following two intervals to be I 1 1, I 1 2 where the superscript means the step and the subscript means the enumeration of the interval within the set. Thus, at step n, there are 2 n intervals written as each of which has length Now, every x K can be written as I n 1, I n 2,..., I n 2 n I n j = 1 3 n We also make the claim that K has length 0. The sum of the intervals that we remove is which is a geometric progression x = k=1 I k m k n 3 n+1 + And thus, this set has length i=0 ( ) i 2 = 1/3 = 1 = [0, 1] 3 1 2/3 Theorem 29. If f : X Y is continuous, and g : Y Z is continuous, then g f : X Z is also continuous Let U Z to be an open set. Then (g f) 1 (U) = f 1 (g 1 (U)) Now, we know that g 1 (U) is open, and thus f 1 (g 1 (U)) is also open which completes the proof. Definition 26. Let X, Y be metric spaces. The map f : X Y is called an Isometry if for all x 1, x 2 X, we have d X (x 1, x 2 ) = d Y (f(x 1 ), f(x 2 )) Example 11. Parseval s Identity If we take f C([0, 2π]) and we define a 0 = 0, a n = 1 2π 2π 0 f(x) sin(nx)dx then 2π 0 b n = 1 2π 2π 0 f(x) cos(nx)dx ( f(x) 2 dx = f 2 2 = C a 2 n + n=0 n=1 f {a = (a 0,..., a n,...), b = (b 1..., b n,...)} b 2 n ) 19

20 or we could write or for and so Thus, we have 1 2π 2π 0 f(x)[cos(nx) + i sin(nx)]dx 1 2π 2π 0 f(x)e inx dx f {(a 0, a 1 + ib 1,..., a n + ib n,...} f 2 2 = C ( a 2 l 2 + b 2 l 2) (a, b = f L 2 = a b 2 2 C Definition 27. (i) f is bijective. Let X, Y be topological spaces. The map f : X Y is a Homeomorphism if (ii) Both f : X Y and f 1 : Y X are continuous funcitons. We say that X, Y are Homeomorphic if and only if there exists a homeomorphism f : X Y. Then open/closed sets, closure, limit points and boundary are all the same for X and Y. Also, continuous functions on X, Y are the same. 20

21 Chapter 3 Completeness Basics & Definitions Section 3.1 Definition 28. A sequence (x n ) in a metric space X is a Cauchy Sequence if for any ɛ > 0, there exists N N such that if m, n > N, then d(x n, x m ) < ɛ Definition 29. We say that a metric space X is Complete if and only if every Cauchy sequence is convergent. That is, if (x n ) is Cauchy, then there exists z X such that as n. Example 12. d(z, x n ) 0 Examples of Complete Metric Spaces The space l 2. Now, l 2 is the space of sequences. Let x 1,..., x k,... l 2 where and such that x k = (x k 1,..., x k n,...) l 2 (x k j ) 2 < Now, suppose that the sequence (x k ) is Cauchy in l 2 which is equivalent to x n x m 2 0 as m, n. We want to find y = (y 1,..., y n,...) l 2 such that x k y as k. We know that x k x l 2 2 = (x k j x l j) 2 (x k m x l m) 2 For each m, (x k m) is cauchy in R. Now there exists y m R such that x k m y m as k. The sequence y j is forced upon us. So, let y = (y 1,..., y n,...) and we claim now that y l 2. We let ɛ > 0 and let N 1 be such that x n x m 2 2 < ɛ if m, n > N 1, and we get N 2 (x m j x n j ) 2 = (x m j x n j ) 2 + (x m j x n j ) 2 j=n

22 and the first sum is ɛ and the second sum is ɛ. Fix n, and let m. As m, x m j y j. This implies that N 2 (y j x n j ) 2 ɛ M 2 is any natural number. let M 2. This gives (y j x n j ) < Now, if (x n ) l 2, we have that d 2 (y, x n ) ɛ and thus y l 2. Now (x y 2 j x 2 j + n j b j) 2 ɛ and so which follows from the inequality lim n (x n j b j ) 2 0 (x n j y j ) 2 ɛ provided that n > M 1 and the fact that ɛ is arbitrary yields our claim. Excercise. Show that l p is complete. Let s examine C([a, b]) with d distance d (f, g) = f g = max f(x) g(x) x [a,b] Theorem 30. The space (C([a, b]), d ) is complete. Let f 1 (x),..., f n (x),... be a Cauchy sequence in C([a, b]). Fix x 0 [a, b]. Then (f n (x 0 )) is a Cauchy sequence. f i (x 0 ) f j (x 0 ) max i(x) f j (x) 0 x [a,b] as i, j simultaneously approach. Therefore, (f n (x 0 )) converges to a limit that we call g(x 0 ). It is clear that d (f i (x), g(x)) 0 as i. Let ɛ > 0, let N N be such that for n, m > N. d (f n, f m ) < ɛ = x [a, b], d(f n (x), f m (x)) < ɛ Let m, then f m (x) g(x). Now, passing to the limit, we get that = f n (x) g(x) ɛ if n > N, then d (f n, g) ɛ 22

23 It now remains to show that g(x) is continuous. Fix ɛ > 0 and choose N > 0 such that d (f n, f m ) < ɛ 2 for each m N. Now, let f N (x) be uniformly continuous on [a, b] (we can do this since [a, b] is a compact subset of R). Let δ > 0 be such that if x, y [a, b] and x y < δ, then f(x) f(y) < ɛ 2 Let m > N. Suppose that f m (x) g(x). As before, let x y < δ, then g(y) g(x) g(y) f N (y) + f N (y) f N (x) + f N (x) g(x) < 3ɛ 2 Theorem 31. Let 1 p <. Then (C([a, b]), d p ) is not complete. We shall define a Cauchy sequence { 0, x [ 1, 0] f n (x) g(x) = 1, x (0, 1] Let Now, 1/n 0 f n (t)dt = { nt, t [ ] 0, 1 f n (t) = n 1, t > 1 1 n 1/n 0 [ nt 2 nt dt = 2 ] 1/n as n. We prove now for p = 1. Let 0, x [ ] 1, 1 2n f n (x) = 2 ( ) [ n x 1 2n, x 1 2, 1 n+1 1, x [ 1 2, 1 ] n so then, for m > n Now, as m, n. 0 = 1 2n 0 2 n ] 0, x 1 2 m+1 2 ( ) [ m x 1 f m (x) f n (x) = 2, x 1 1 m+1 2, m+1 1, x [ 1 1 2, m 1 2 ( ) [ n x 1 2, x 1 n+1 2, 1 n+1 d 1 (f m, f n ) = 1/2 n 1/2 m+1 f m (x) f n (x) dx ] 2 m ] 2 n+1 ] 2 n ( 1 2 n 1 ) 2 m Exercise. Let 0, x [ 1, 0] h n,p (x) = (nx) 1/p, x [ ] 0, 1 n 1, x [ 1 n, 1] 23

24 and [ 1/p [ 1/n 1/p [ 1/n 1/p 1/n h n,p (x) dx] p = ( nx 1/p ) dx] p = (nx)dx] = 0 as n. Use h n,p (x) to modify the construction for p = ( ) 1/p 1 0 2n Theorem 32. The space X is complete if and only if for every sequence of nested closed balls B(x n, r n ) B(x 1, r 1 ) such that r n 0 as n. We then have a nonempty intersection. (= ) Let (x n ) be a sequence of centres of the balls. Then, (x n ) is a Cauchy sequence. Indeed, d(x n, x m ) max(r n, r m ) 0 as n, m. X is complete, so x n y X as n. We want to show that We know that for x m B(x n, r n ), m > n, we get y 1 B(x i, r i ) i=n y = lim n x n lim n d(x m, x n ) r n = d(y, x n ) r n ( =) We must prove now that if X is not complete, then a sequence of balls such that r n 0 and B(x n, r n ) B(x 1, r 1 ) B(x n, r n ) = n=1 Suppose now that X is not complete. Then there exists a Cauchy sequence (x n ) such that x n doesn t have a limit in X. Also, there exists n 1 such that d(x m, x n1 ) < 1 2 for m n 1. Now, let B 1 = B(x n1, 1), then there exists n 2 > n 1 such that d(x m, x n2 ) < = 1 4 for each m n 2. Now let B 2 = B(x n2, 1/2). We claim that B 2 B 1. The induction step is that there exists n k > n k 1 such that d(x m, x nk ) < 1 2 k for all m n k. Now, let B k = B (x nk, 12 ) k 24

25 Now, we claim that B k B k 1. We know that where B 2 B 1 r k = 1 2 k 0 as k. Any point lying in must be a limit of (x k ). We assumed that (x k ) doesn t converge. Thus B k = k=1 k=1 B k and this contradiction completes the ( =) direction of the proof. In general, there exist complete metric spaces, and there exists r n that B(x 2, r 2 ) B(x 1, r 1 ) but that B(x k, r k ) = k=1 Hint (Problem 5(ii), Assignment 1): Prove that that doesn t converge to zero such and then a/3 n is dense in [0, 2] where 0 a 2 3 n. {All 3-adic rationals in [0, 2]} K + K Completion & Density Revisited Section 3.2 Definition 30. The Completion of an incomplete metric space X is a metric space Y such that f : X Y is a function that satisfies f is 1-to-1, with [f is an isometry from X to f(x)] f(x) as dense in Y Y is complete d(f(x 1 ), f(x 2 )) = d(x 1, x 2 ) Proposition 33. Every incomplete metric space has a completion that is unique up to isometry. Proof (Idea). Let Z be the set of all Cauchy Sequences x = (x 1,..., x n,...) in X. 25

26 Definition 31. as n. We say that two Cauchy sequences (x n ) and (y n ) are Equivalent if and only if d(x n, y n ) 0 Proposition 34. Equivalence is an equivalence relation. The only non-trivial part of this part is transitivity. Thus, if (x n ) (y n ) and (y n ) (z n ), then we know that for n > n 1, d(x n, y n ) < ɛ 2 and for n > n 2, d(y n, z n ) < ɛ 2 and thus for n > N = max{n 1, n 2 }. d(x n, z n ) < ɛ Definition 32. If we let Y denote the set of all equivalence classes in Z with respect to, then for (x n ), (y n ) Cauchy sequences in X, we define ( ) d (x n ), (y n ) = lim d X(x n, y n ) n Proposition 35. So, if (x n ) (x n), then We have d = 0 (x n ) (y n ) lim d(x n, y n ) = lim n n d(x n, y n ) f : X Y such that f(x) = {x, x, x,...}. It remains to show that f(x) is dense in Y Y is complete For the first part, let (x 1 n), (x 2 n),..., (x k n) be Caucy sequences. We know that d Y ((x k1 n ), (x k2 n )) 0 as k 1, k 2 infty. Now, let y Y. y is an equivalence class of the sequence (x n ). Let ɛ > 0, and let N be such that d(x n, x m ) < ɛ for all n, m > N. Choose, n > N, and we get d Y (y, f(x n )) = d Y (y, (x n, x n,..., x n,...)) ɛ where f(x n ) f(x). This implies that f(x) is dense in Y. For the second part, let y 1,..., y n be Cauchy in Y. Choose { x n : d X (f(x n ), y n ) < 1 } n We claim that (x n ) is a Cauchy sequence in X. If z = (x n ), then y n z in Y. 26

27 We now ask the question: Why are complete metric spaces cool? A possible answer would involve the Contraction Mapping Principle. O.D.E. & Contraction Section 3.3 Definition 33. If X is a metric space and A : X X, then we say that A is a Contraction Mapping if there exists 0 < α < 1 such that d(a(x), A(y)) < α d(x, y) for any x, y X. It is a fact that if A is a contraction mapping, then A is continuous. Theorem 36 (Contraction Mapping Principle). If X is complete, and A is a contraction mapping, then there exists a unique point x 0 X such that A(x 0 ) = x 0 is a fixed point of A. Moreover, for any x X, A n (x) x 0 as n. Let x X and consider {x, A(x),..., A n (x),...} = {x 0,..., x n,...} We claim that (x n ) is a Cauchy sequence. To this end, we begin with By induction on m, we see d(a m (x), A m+k (x)) = d(a m (x), A m (A k (x))) α m d(x, A k (x)) d(a 2 (x), A 2 (y)) αd(a(x), A(y)) α 2 d(x, y) and so if m, then α m 0. Now, we know that A n (x) x 0 as n, since X is complete. d(x, A k (x)) for each k. d(x, A(x)) = b Where and so d(x, A k (x)) d(x, A(x)) + d(a(x), A 2 (x)) + + d(a k 1 (x), A k (x)) which is the bound we need. We also know that d(a i (x), A i 1 (x)) α i b d(x, A k (x)) b(1 + α + + α k 1 ) A n+1 (x) A(A n (x)) A(x 0 ) We now want an a priori bound on b 1 α as n. And thus A(x 0 ) = x 0 which means that x 0 is a fixed point of A. A cannot have two fixed points since if x 0, y 0 are fixd, then Ax 0 = x 0 A(y 0 ) = y 0 then which is a contradiction that completes the proof. d(a(x 0 ), A(y 0 )) = d(x 0, y 0 ) 27

28 Remark. Alternatively to our original proof, we let our sequence be x n = Ax n 1 so that x n = A n x 0 and since A is continuous, and since x n x, we get that which shows existence of a fixed point. Ax = A lim n x n = lim n Ax n = x Definition 34. We say that a map f is Lipschitz, denoted f Lip K if f(x) f(y) < K x y and if K < 1 then f is a contraction mapping. Example Suppose that f : [a, b] R and that f is Lipschitz with K > 0. If K < 1 then we know that f is a contraction mapping from [a, b] [a, b]. Hence if x 0, x 1 = f(x 1 ), x 2 = f(f(x 0 )),... then f n (x) y such that f(y) = y. Sufficient condition is that if f C 1 ([a, b]), f (t) K < 1 for all t [a, b]. for some t [x, y] [a, b]. So f(y) f(x) = (y x) f (t) f(y) f(x) y x = f (t) K < 1 2. Solving ODE s. Now, suppose that d dx y = f(x, y) y(x 0) = y 0 f(x, y 1 ) f(x, y 2 ) M y 1 y 2 for M fixed, then f Lip M in y. We assume that f is continuous on some rectangle R R 2 such that (x 0, y 0 ) R. Then, on some small interval x x 0 d there exists a unique solution to the ODE satisfying the initial condition y(x 0 ) = y 0. Solving the ODE above is equivalent to solving an integral equation and we note that f is continuous on R and so ϕ(x) = y 0 + x x 0 f(t, ϕ(t)) dt ϕ(x 0 ) = y 0 ϕ (x) = f(x, ϕ(x)) f(x, y) K for all points (x, y) R 1 where R 1 contains (x 0, y 0 ). Now we choose d > 0 so that if x x 0 d and y y 0 Kd, and 28

29 Md < 1 then (x, y) R 1. Let C be the space of continuous functions ϕ on [x 0 d, x 0 + d] such that Let d be the sup distance d(ϕ 1, ϕ 2 ) = x y 0 K d sup ϕ 1 (x) ϕ 2 (x) x [x 0 d,x 0+d] then C is complete. Now we need to define a contraction mapping and so let Aϕ = ψ(x) = y 0 + x x 0 f(t, ϕ(t)) dt So we claim that A is a contraction map from C C. If this claim is true, then there exists a unique fixed point ϕ of A such that Aϕ = ϕ which means that this unique ϕ solves the integral equation uniquely and thus solves the ODE uniquely. Now, to prove the claim, we define a starter function ϕ C such that and we have that x x 0 d x ψ(x) y 0 = A(ϕ(x)) y 0 = f(ϕ(t), t) dt x 0 max f(ϕ(t), t) x x 0 t [x 0,x] Kd Now, for the contraction (with respect to the sup norm) x x A(ϕ 1 (x)) A(ϕ 2 (x)) = y 0 + f(t, ϕ 1 (t)) dt y 0 f(t, ϕ 2 (t)) dt x 0 x 0 x = [f(t, ϕ 1 (t)) f(t, ϕ 2 (t))] dt x 0 Thus, remember that M d < 1 and M = α. x x 0 f(t, ϕ 1 (t)) f(t, ϕ 2 (t)) dt x M ϕ 1 (t) ϕ 2 (t) dt x 0 M x x 0 max ϕ 1(t) ϕ 2 (t) t [x 0,x] M d d (ϕ 1, ϕ 2 ) d (A(ϕ 1 ), A(ϕ 2 )) M d d (ϕ 1, ϕ 2 ) Fact. Let X be a complete metric space. If A n is a contraction map from X. Then A(x) = x also has a unique solution. We will leave the proof of this until when we need to use it. 29

30 Chapter 4 Compactness Basics & Definitions Section 4.1 Definition 35. Let A X where X is a metric space. We say that A is Sequentially Compact if every sequence in A has a subsequence which converges to some x X (x / A is possible). Definition 36. A X is called an ɛ-net if for each x X, there exists y A such that d(x, y) ɛ Definition 37. X is Totally Bounded if for each ɛ > 0, there exists a finite ɛ-net in X. OR X is Totally Bounded if for any ɛ > 0, there exists a finite set x 1,..., x n such that where n = n(ɛ). X B(x 1, ɛ) B(x 2, ɛ) B(x n, ɛ) Other problems where knowing what n(ɛ) is useful include Coding Complexity We note first that a totally bounded space is bounded. That is, d(x 1, x 2 ) 2ɛ + max d(y 1, y 2 ) = 2ɛ + diam(ɛ net A) y 1,y 2 ɛ net A Fact. If B X is totally bounded, then B X is also totally bounded. We now ask the question: What is the minimal number of points in an ɛ-net? [0, 1] n, least number of points ( ) n 1 ɛ + 1 Hint: We can use without proof the fact that all norms in R 2 are equivalent. This implies that we can use our favourite norm in R n to define the operator norm, namely Ax p x p or Ax x 30

31 Now, suppose that C 1 < C 2, so if then 1 d 1(x, y) C 1 d 2 (x, y) < C 1 1 d 1(x, y) C 2 d 2 (x, y) < C 2 Example 14. Examples of totally bounded sets. In R n, a set is totally bounded iff the set is bounded (n is fixed). In l 2, take the set A = We claim that A is totally bounded. Let ɛ > 0 and choose m such that Now, let { x = (x 1,..., x n,...) : x 1 1, x 2 1 2, x n 1 } 2 n 1 2 m < ɛ 2 C 1 = {x = (x 1,..., x m, 0, 0,..., 0,...) : x A} First, we claim that C 1 is totally bounded. C 1 can be covered by N(ɛ) < ɛ m balls of radius ɛ. Now we claim that the whole set C lies in an ɛ-neighbourhood of C 1. It suffices to show that for any x C, there exists y C 1 such that d(x, y) ɛ To this end, take x = (x 1,..., x m, x m+1,...) C and y = (x 1,..., x m, 0,..., 0,...) C 1. Now, d 2 (x, y) = k=m+1 = 1 4 m x 2 k 1 (2 m+1 ) ( = m 1 1/4 = (2 m+2 ) 2 + ) m 1 < ɛ2 Now, the ɛ-net in C 1 is also a finite (2ɛ)-net in C. So since, ɛ is arbitrary, we conclude that C is totally bounded. We remark that the same construction would work if x k a k s.t. a 2 k < k=1 We claim that the whole of l 2 is not totally bounded. To this end, define x 1 = (1, 0, 0, 0,...) x 2 = (0, 1, 0, 0,...) x 3 = (0, 0, 1, 0,...) 31

32 We know that A = {x 1, x 2,...} d 2 (x m, x n ) = = 2 for any m, n. So, for any ɛ < 2/2, A cannot be covered by a finite ɛ-net. Theorem 37. Let X be complete, and suppose that A X. A is sequentially compact if and only if A is totally bounded. (= ) Suppose that A is not totally bounded. There exists an ɛ > 0 such that A doesn t have a finite ɛ-net. Choose x 1 A, then there exists x 2 A such that and there exists x 3 such that d(x 3, x 1 ) ɛ d(x 1, x 2 ) ɛ d(x 3, x 2 ) ɛ and continuing this way, we see that there exists x k such that for any j < k, d(x k, x j ) ɛ Now, we claim that x j cannot have a convergent subsequence. To yield this claim, we simply say that any subsequence is not Cauchy. ( =) Suppose that X is complete and A is totally bounded. Let (x a ) be a sequence of points in A. Let ɛ 1, ɛ 2 = 1 2,..., ɛ k = 1 k For any k, there exists a finite set a k 1, a k 2,..., a k n k such that n B(a k i, ɛ k ) = A i=1 One of B(a 1 j, 1) has infinitely many x k s. In a ball B i of radius 1, there is an infinite subsequence of x k s Also, one of the balls x (1) 1,..., x(1) n,... ( B a 2 j, 1 ) 2 has infinitely many x (1) j. Call it B 2. Call the corresponding subsequence x (2) 1,..., x(2) n,... One of ( B a k j, 1 ) k has infinitely many points of x (k 1) 1,..., x (k 1) n,... Call it B k. Elements lying in B k for a subsequence x (k) 1,..., x(k) n,... Now, let y k = x (k) k. y k is a subsequence of x k s. y k is Cauchy. For m 1, ( ( d(y k, y k+m ) diam B a k j, 1 )) 2 k k X is complete. y k z X. 32

33 Arzela & Compacta Section 4.2 Proposition 38. X is complete, and A X is compact if and only if for all ɛ > 0, there exists in X a sequentially compact ɛ-net that covers A. Definition 38. Let F = {ϕ(x)} be a family (collection) of functions on [a, b]. We say that F is Uniformly Bounded if there exists M > 0 such that ϕ(x) M, x [a, b], ϕ F Definition 39. We say that the family F is Equicontinuous if for every ɛ > 0, there exists δ > 0 such that x 1, x 2 [a, b], with x 1 x 2 < δ, and ϕ F, we get ϕ(x 1 ) ϕ(x 2 ) < ɛ Remark. Both of the above definitions work for any metric space X. Theorem 39 (Arzela). Let F be a family of continuous functions on [a, b]. Then F is sequentially compact in (C([a, b]), d ) (which is complete by Theorem 30) if and only if F is uniformly bounded and equicontinuous. (= ) It is true that F is sequentially compact in (C([a, b]), d ) if and only if F is totally bounded in C([a, b]) by Theorem 36. Now, let ɛ > 0, then F can be covered by a finite (ɛ/3)-net. There exists ϕ 1,..., ϕ k C([a, b]) such that for any ϕ C([a, b]), we have d (ϕ, ϕ j ) < ɛ 3 for some 1 j k. Now, ϕ j (x) M j for all x [a, b]. Take M = max{m j } + ɛ 3 and for any x [a, b] and for any ϕ F, there exists 1 j k such that ϕ(x) ϕ j (x) ɛ 3 Now, for all x, ϕ(x) ϕ j (x) + ɛ 3 M j + ɛ 3 M and since d (ϕ, ϕ j ) ɛ 3 which means that F is uniformly bounded. Each ϕ(x) C([a, b]) is uniformly continuous, so there exists δ j such that x 1 x 2 < δ j = ϕ j (x 1 ) ϕ j (x 2 ) < ɛ 3 and let δ = min{δ j }. Let x 1, x 2 [a, b] such that x 1 x 2 < δ. Let ϕ F and choose 1 j k so that d (ϕ, ϕ j ) < ɛ 3 33

34 and now ϕ(x 1 ) ϕ(x 2 ) ϕ(x 2 ) ϕ j (x 1 ) + ϕ j (x 1 ) ϕ j (x 2 ) + ϕ j (x 2 ) ϕ(x 2 ) ɛ 3 + ɛ 3 + ɛ 3 = ɛ ( =) We know that (C([a, b]), d ) is complete and hence it is sequentially compact if and only if it is totally bounded. We claim that F is totally bounded. Let ɛ > 0. By uniform boundedness, we get that ϕ(x) M Now, let δ > 0, then for any x 1, x 2 [a, b], we have that for any ϕ F. Now, let and let x 0 = a x 1 = a + b a N x 1 x 2 < δ = ϕ(x 1 ) ϕ(x 2 ) < ɛ 5 b a N x 2 = a + 2 b a N < δ x N 1 = a + (N 1) b a N x N = b Now we let m > 0 and 2M m < ɛ 5 and y 0 = M and y m = M so that the points y k subdivide the interval [ M, M] into m equal parts. Now, we take any continuous function ϕ F. We look at values {ϕ(x 0 ), ϕ(x 1 ),..., ϕ(x N )} and for each j, we choose y j such that ϕ(x j ) y j ɛ 5 Now let ψ be a piecewise-linear function such that ψ(x j ) = y j. The set of all possible ψ (call it A) is finite and A (M + 1) n+1 Furthermore, since we know that the slope is between 3 and 3, it follows that the number of functions A (M + 1) 7 n Now we claim that the set A of possible ψ form an ɛ-net in F. 0 k n, To this end, we first note that for any ϕ(x k ) ψ(x k ) < ɛ 5 and ϕ(x k+1 ) ψ(x k+1 ) < ɛ 5 by our choice of ψ. Also, by uniform continuity, we get ϕ(x k ) ϕ(x k+1 ) < ɛ 5 = ψ(x k ) ψ(x k+1 ) < 3ɛ 5 Now, for any x [x k, x k+1 ], we have by linearity that ψ(x) ϕ(x k ) 3ɛ 5 34

35 Then ϕ(x) ψ(x) ϕ(x) ϕ(x k ) + ϕ(x k ) ψ(x k ) + ψ(x k ) ψ(x) = α + β + γ We know that α < ɛ/5 by uniform continuity, and β < ɛ/5 by construction of ψ and γ < 3ɛ/5. Thus ϕ(x) ψ(x) < ɛ for any x [a, b] which completes the proof.. Recall that A X is sequentially compact if and only if any sequence (x n ) A has a subsequence that converges in X. Definition 40. A subset A X is Sequentially Compact In Itself if and only if any sequence (x n ) A has a convergent subsequence that converges in A. Example 15. Example 16. Definition 41. If B = Q [0, 1] then B is sequentially compact but not in itself. If A = X then sequential compactness is equivalent to sequential compactness in itself. A sequentially compact metric space is called Compactum. Proposition 40. Let A X. If A is sequentially compact, then A is sequentially compact in itself if and only if A is as a closed subset of X. Corollary 41. Proposition 42. bounded. EXCERCISE Proposition 43. Any closed bounded subset of R n is sequentially compact in itself. Let X be a metric space. X is a compactum if and only if X is complete and totally Every compactum has a countable dense subset. Theorem 44. The following are equivalent: (i) X is a compactum. (ii) An arbitrary open cover {U α } α I of X has a finite subcover. That is there exists α 1,..., α n I such that n X (iii) (Finite Intersection Property) A family {F } α I of closed subsets of X such that every finite collection of F n s has a nonempty intersection has F α α I ((i) (ii)) Suppose that X is sequentially compact in itself. Let ɛ n = 1/n and take a finite ɛ n -net; with i=1 U αi 35

36 centres a (n) k for each n so that X ( B a (n) k, 1 ) n We proceed by contradiction. Assume that there exists an open cover {U α } without a finite subcover and choose one of ( B a (n) k, 1 ) n that cannot be covered by a finitely many U α s. Now, say ( B a (n) k, 1 ) 0(n) n Let x n = {x (n) k }. Now, X is a compactum, so a subsequence x 0(n) n j y X. y U β, for some β I. U β is open and so B(y, ɛ) O β for some ɛ > 0. Now choose n so that and then Now, we claim that 1 n < ɛ 2 d(y, a (n) k 0(n) ) < ɛ 2 ( B a (n) k, 1 ) 0(n) B(y, ɛ) O β n which is trivially true. But this is a contradiction which completes this part of the proof. ((ii) = (i)) Suppose that any open cover of X has a finite subcover. Then we claim that X is complete, and that X is totally bounded. To show total boundedness, we let ɛ > 0 and then X B(x, ɛ = U x ) x X has a finite subcover. Thus, there exists x 1,..., x n(ɛ) such that X n(ɛ) k=1 B(x k, ɛ) is a finite ɛ-net and ɛ is arbitrary, so total boundedness follows. To show completeness, we let B k B 1 be a sequence of closed, nested spheres of radius r k 0 as k. Now, suppose, for a contradiction, that Now, B k = k=1 (X\B k ) k N is an open cover of X. It cannot have a finite subcover, otherwise B j = for j > N which is a contradiction that completes the final portion of the proof. 36

37 Remark. Open cover property is usually taken as a definition of compactness for general topological spaces. To get a sequential definition for general topological spaces, one should generalize the notion of sequence to nets. Theorem 45. Let If X is compact, and f : X Y is continuous, then f(x) is compact. f(x) α I V α Then, for any α, define f 1 (V α ) = U α which is open and {U α } α I is an open cover of X which is compact. Then there exists a finite subcover X U α1 U α2 U αn which implies that is a finite subcover of f(x) which yields our claim. f(x) V α1 V αn Theorem 46. If X is compact and f : X Y is continuous, one-to-one and onto, then f 1 is also continuous and hence f is also a homeomorphism. Define Y = f(x) to be compact. Let B X be closed. Then B is compact since closed subsets of compact metric spaces are compact. Now, f(b) = C is closed in Y which completes the proof. If we let X, Y be compact metric spaces and if C(X, Y ) is the set of continuous functions from X to Y. Now, let f, g C(X, Y ) and define d (f, g) = sup d Y (f(x), g(x)) x X Theorem 47. Let D C(X, Y ) where X, Y are compact. Then D is compact in C(X, Y ) if and only if for every ɛ > 0, there exists δ > 0 such that for every x 1, x 2 X, d X (x 1, x 2 ) < δ = d Y (f(x 1 ), f(x 2 )) < ɛ for every f D. (Sketch) Let M X,Y be the space of all mappings (not necessarily continuous) from X to Y. Define d as before. Now, C(X, Y ) is a closed subset of M X,Y. Also, if f n : X Y converges uniformly with respect to d then the limit function is continuous. Theorem 48. A closed subset Y of a compact set X is compact. Let (U α ) be a cover of Y. (U α ) (x\y ) is an open cover of X which is compact. This implies that a finite subcover U α1,... U αk, (X\Y ) possibly of X. Hence, the subcover covers Y. Proposition 49. compact in itself. Let X be a metric space. Then Y is sequentially compact in X if and only if Y is Proposition 50. Any function that is continuous on a compact metric space X is uniformly continuous. Suppse, for a contradiction that there exists ɛ > 0 such that there exists (x n ), (x n) in X such that d(x n, x n) < 1 n = f(x n) f(x n) ɛ 37

38 Now, X is compact, so there exists x nk y X. Then x n k y. f is continuous and so f(x nk ), f(x n k ) f(y). This contradicts our assumption and the proof is complete. Proposition 51. If X is compact, then f : X R attains a least upper bound U and greatest lower bound L. Suppose, for a contradiction, that U is not attained. Then, for each n, there exists x n X such that U > f(x n ) U 1 n by sequential compactness, x nk y X. Now, by continuity, f(x nk ) f(y). Thus, f(y) = U which contradicts our assumption and completes the proof for attaining the supremum. For the infimum, it is symmetric. We just replace f by f. Theorem 52. (General Arzela) Let D C(X, Y ) where X, Y are compact metric spaces. Then D is sequentially compact in C(X, Y ) (which is equivalent to D is compact in C(X, Y )) if and only if for every ɛ > 0, there exists δ > 0 such that for every x 1, x 2 X, d(x 1, x 2 ) < δ = d Y (f(x 1 ), f(x 2 )) < ɛ for every f D. We remarked that it suffices to show that it suffices to show that D is totally bounded in M(X, Y ) with respect to d (f, g) = sup d Y (f(x), g(x)) x X where M(X, Y ) is the space of all maps from X to Y (not necessarily continuous). Also, it was proven that C(X, Y ) is closed in M(X, Y ) To prove that D is totally bounded, we shall be approximating functions in D by piecewise constant (but discontinuous) functions. Now, let ɛ > 0, and in the definition of equicontinuity, choose δ > 0 such that d X (x 1, x 2 ) < δ = d Y (f(x 1 ), f(x 2 )) < ɛ for all f D. Let {x 1,..., x n } be a (δ/2)-net in X. Then let ( A 1 = B x 1, δ ) ( ; A 2 = B x 2, δ 2 2 ) \A 1 ; A 3 = B ( x 3, δ ) \(A 1 A 2 ) 2 so that ( A k = B x k, δ ) \(A 1 A k 1 ) 2 Now, A 1 A n = X. The A j are disjoint. Also, if x 1, x 2 A i, then d X (x 1, x 2 ) < δ Now, if Y is compact, then there exists a finite (ɛ/2)-net {y 1,..., y m } Y. We shall approximate functions in D by the set of mappings that are constant on A j s and take values in {y 1,..., y m }. This is a finite set and the number of functions is less than or equal to m n. We call this set Φ and we claim that for every f D, there exists ϕ Φ such that d (f, ϕ) 2ɛ. To this end, for all 1 i n, there exists 1 j m such that d Y (f(x i ), y j ) < ɛ 38

39 Now, let ϕ Φ be defined by letting ϕ(x i ) = y j, then d Y (f(x), ϕ(x)) d Y (f(x), f(x i )) + d Y (f(x i ), ϕ(x i )) + d Y (ϕ(x), ϕ(x i )) The first two terms are less than ɛ and the last term is 0. Thus, d Y (f(x), ϕ(x)) < 2ɛ and the proof is complete. Example 17. Let f n (x) = x n and we wish to show that {f n } C([0, 1]) is uniformly bounded and uniformly equicontinuous. Clearly, it is uniformly bounded by 1. For uniform equicontinuity, we have and so f n(x) = nx n 1 f n(1) = n We suppose for a contradiction that for every ɛ > 0, there exists δ > 0 such that for each x, y X, for each n N. By Taylor s theorem, we get the rest is provided in a note online. x y < δ = f n (x) f n (y) < ɛ f n (x) f n (y) = f n(x) x y 39

40 Chapter 5 Basic Point-Set Topology Product Topology Section 5.1 The basis of open sets are the sets where (i) U j s are open in X j s (ii) U j = X j except for finitely many j s Proposition 53. Let U 1 U n f : Y X = be a function. Then f is continuous (with respect to the product topology) if and only if i=1 X i f(y) = [f 1 (y),, f n (y), ] where the f i (y) are continuous for each i. It suffices to show that f 1 (Basis element of the product topology) is open in Y. Let and so U = U 1 U n X n+1 X n+2 f 1 (U) = f1 1 (U 1) fn 1 (U n ) Y Y = V is open, as we disregard the intersections with Y and also since each f 1 i (U i ) is open. Example 18. be such that If we let f : R R x (x,, x) then we can see that f is not continuous in the box topology. Theorem 54. (Easy Version Of Tikhov Theorem) Suppose that X j is compact for each j, then X = i=1 with the product topology is also compact. Suppose that X has an open cover O that has no finite subcover. We first claim that there exists x 1 X 1 such that no basis set of the form U 1 U 2 X i 40

41 is overed by finitely many open sets O. To show that this claim holds, we will suppose, for a contradiction that for any x 1, there exists U 1 containing x 1 such that is covered by finitely many sets in O. Then the sets are open covers of X. If X 1 is compact, then U 1 (x 1 ) X 2 {U(x 1 ) : x 1 X 1 } X 1 U 1 (x 1 ) U 1 (x k ) and each U j X 2 X n is covered by finitely many open sets in O. Thus X 1 X n is covered by finitely many sets in O which contradicts our assumption proving the claim. Now, by induction, there exists X 2 X 2 such that no basis set of the form U 1 U 2 X 3 X n such that (x 1, x 2 ) U 1 U 2 can be covered by finitely many open sets in O. This is step 2 in the induction. This is proven using the compactness of X 2. Now, for step k, we have that for each k N, there exists x k X k such that no element of the form U 1 U k X k+1 can be covered by finitely many open sets in O. Now, consider x = (x 1,, x n, ) X 1 X k and so there exists v O, such that x V. So there exists a basis element U = U 1 U m X m+1 which contradicts step m of the induction. Suppose now that X is a metric space. Is the product topology metrizable? The answer is yes. We wish to put a metric on Step 1. Let d j be the metric on X j. Replace d j by X j d j (x, y) = d j(x, y) 1 + d j (x, y) 1 It is a fact that d j preserves topology on X j. Step 2. Let x = (x 1,, x n, ) y = (y 1,, y n, ) and we say that and we re done! D(x, y) = k=1 d j (x k, y k ) 2 k 41

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